Geometric Bites

Why the Line ax + by = 0 Passes Through the Point (−b, a) In ℝ² , the equation ax + by = 0 describes a line that is perpendicular to the vector (a, b) . This article explains exactly why—and why that line always passes through the point (−b, a) . 1. Start with the Vector (a, b) Consider the vector (a, b) . To find a line perpendicular to it, we need a vector whose dot product with (a, b) is zero. Try the vector (−b, a) : (a, b) · (−b, a) = a(−b) + b(a) = −ab + ab = 0 Therefore, (−b, a) is perpendicular to (a, b) . 2. Any Scalar Multiple Also Works If (−b, a) is perpendicular to (a, b) , then any multiple λ(−b, a) is also perpendicular: (a, b) · [λ(−b, a)] = λ[(a, b) · (−b, a)] = λ · 0 = 0 Let this perpendicular vector be (x, y) . Then (x, y) = λ(−b, a) Every point on the line comes from a particular choice of λ . 3. Converting to an Equation Since (x, y) is perpendicular to (a, b) , we have: (a, b) · (x, y) = 0 Expanding ...

Converting the Vector Equation of a Line into Cartesian Form A straight line in three-dimensional space can be expressed using vectors. One important vector form is (𝐑 − 𝐀) × 𝐁 = 0 This equation states that the displacement vector from a fixed point 𝐀 to a general point 𝐑 is parallel to the direction vector 𝐁. Two non-zero vectors have a zero cross product precisely when they are parallel. From this fact, the Cartesian (symmetric) equation of the line can be derived. 1. Substituting Coordinate Vectors The general point on the line is written as 𝐑 = (x, y, z) The fixed point is 𝐀 = (x₁, y₁, z₁) The direction vector is 𝐁 = (l, m, n) Substituting these into the vector equation yields: ((x, y, z) − (x₁, y₁, z₁)) × (l, m, n) = 0 which simplifies to: (x − x₁, y − y₁, z − z₁) × (l, m, n) = 0 2. Using the Condition for a Zero Cross Product If two non-zero vectors have a zero cross product, then one is a scalar multiple of the other. T...

Normalised Vectors: A Clear and Intuitive Guide Vectors can have any length, but many mathematical problems only depend on direction. To separate direction from magnitude, we normalise the vector. This produces a new vector of length 1 that points the same way as the original. Normalised vectors are central to geometry, physics, 3D graphics, transformations, and any setting where orientation matters. By working with a unit vector, calculations become simpler, cleaner, and more meaningful. What Is a Normalised Vector? A normalised vector is a vector with magnitude 1. It keeps its direction but loses its original size. Some simple unit vectors include: (1, 0, 0) — magnitude 1 (0, 1, 0) — magnitude 1 (0, 0, 1) — magnitude 1 These are the standard basis vectors. In general, any non-zero vector can be transformed into a unit vector by dividing by its magnitude. Normalising a Vector in 2 Dimensions For a 2D vector (a, b) , the length is: √(a² + b²) To n...

The Dot Product Identity and the Cosine Rule in ℝ 3 In this article we derive the dot product identity A · B = |A| × |B| × cos(θ) and show how this identity leads directly to the cosine rule, using a combination of coordinate algebra and geometric interpretation. 1. Vectors in ℝ 3 Let the vectors be: A = (a 1 , a 2 , a 3 ) B = (b 1 , b 2 , b 3 ) Their difference is: A - B = (a 1 - b 1 , a 2 - b 2 , a 3 - b 3 ) The squared magnitude of this difference vector is: |A - B| 2 = (a 1 - b 1 ) 2 + (a 2 - b 2 ) 2 + (a 3 - b 3 ) 2 . 2. Expanding the Square of the Difference Expand each component: (a 1 - b 1 ) 2 = a 1 2 - 2a 1 b 1 + b 1 2 (a 2 - b 2 ) 2 = a 2 2 - 2a 2 b 2 + b 2 2 (a 3 - b 3 ) 2 = a 3 2 - 2a 3 b 3 + b 3 2 Adding these three expansions gives: |A - B| 2 = (a 1 2 + a 2 2 + a 3 2 ) + (b 1 2 + b 2 2 + b 3 2 ) - 2(a 1 b 1 + a 2 b 2 + a 3 b 3 ). Recognise the squared magnitudes: |A| 2 = a 1 2 + a 2 2 ...

🧭 A Primer for Cross Product Calculations Right Hand Rule for Cross Product. Source: https://commons.wikimedia.org/wiki/File:Right-hand_rule_for_cross_product.png 1. Basis Vectors î = (1, 0, 0), ĵ = (0, 1, 0), k̂ = (0, 0, 1) These are unit and mutually perpendicular vectors. 2. Dot Product (for reference) 𝐀 · 𝐁 = |𝐀| |𝐁| cos θ î · ĵ = ĵ · k̂ = k̂ · î = 0 î² = ĵ² = k̂² = 1 3. Definition of the Cross Product 𝐀 × 𝐁 = |𝐀| |𝐁| sin θ n̂ θ is the angle from 𝐀 to 𝐁. n̂ is a unit vector perpendicular to both 𝐀 and 𝐁. The direction of n̂ follows the right-hand rule (anticlockwise = positive, clockwise = negative). 4. Fundamental Basis Cross Products î × ĵ = k̂ ĵ × k̂ = î k̂ × î = ĵ Reversing the order changes the sign: ĵ × î = −k̂ k̂ × ĵ = −î î × k̂ = −ĵ And any vector crossed with itself is zero: î × î = ĵ × ĵ = k̂ × k̂ = 0 5. Expansion in Component Form 𝐀 = a₁ î + a₂ ĵ + a₃ k̂ 𝐁 = b₁ î + ...